I am trying to understand the derivation of the formula for the sum of an arithmetic sequence of the first $n$ terms.
I do not understand what rules or reasoning allow two sequences to be added in reverse order to eliminate the common difference $d$ and arrive at the conclusion that the sum of an arithmetic sequence of the first $n$ terms is one half $n$ times the sum of the first and last terms. This seems to be a contrived way to eliminate the common difference from the expanded based on some unexplained knowledge of $d$ and arithmetic sequences in general.
I have researched this question in maths textbooks and online and each time the derivation is presented I cannot seem to find an explanation as to why it would be evident to a mathematician that by adding the sequences they would derive the formula.
The background.
The derivation of the formula as explained in many textbooks and online sites is as follows.
- To find the sum of an arithmetic sequence for the first $n$ terms $S_n$, we can write out the sum in relation to the first term $a_1$ and the common difference $d$.
$$ S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + ... + a_n $$
- It is also possible to write the sequence in reverse order in relation to the last term $a_n$.
$$ S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + ... + a_1 $$
- When we add these sequences together we derive the formula for the sum of the first n terms of an arithmetic sequence.
$$ \begin{array}{r} S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + \ldots + a_n \\ + \,S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + \ldots + a_1 \\ \hline 2S_n = (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) \ldots \end{array} $$
- Because there are $n$ many additions of $(a_1 + a_n)$ the lengthy sum is simplified as $n(a_1 + a_n)$ and solving for $S_n$ we arrive at the formula.
$$ S_n = \frac{n}{2}(a_1 + a_n) $$
Unfortunately I can't seem to find the reasoning in any of these explanations as to why the two sequences (ordinary order and reverse) were added. It makes sense to me that they were added but not why this was the next logical step when deriving the formula.
The question.
Why were the two sequences added to derive the formula and what does that show about the nature of arithmetic sequences?
In my attempt to figure this out I noted that by studying many sequences we can see that the ratio of the sum of the sequence for the first $n$ terms $S_n$ and the sum of the first and last terms $(a_1 + a_n)$ is always $\frac{n}{2}$ for any arithmetic sequence. So possibly it could be said by induction that if for any arithmetic sequence it is true that:
$$ \frac{S_n}{a_1 + a_n} = \frac{n}{2} $$
Then it must also be true that:
$$ S_n = \frac{n}{2}(a_1 + a_n) $$
However, to me this still doesn't explain why the derivation decides to add the two sequences.
$\endgroup$ 62 Answers
$\begingroup$Commutativity of addition lets you permute the sum of two addends. By induction on the number of addends you can extend this to any finite number of addends. Associativity then lets you group them however you want.
$\endgroup$ 4 $\begingroup$Think back to the sum of triangle numbers. To add the first n numbers, you take the first and last and combine them to get n+1, then you do the same for the second and second to last to get n+1. For a visual of this process, look at . Here in your question, your triangle is a little lopsided and doesn't come to a point, but the idea remains the same :).
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